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Here it is said that a $2\times 2$ matrix $A$ is positive definite if and only if $tr(A) >0$ and $det(A)>0$. This will not work if $A$ is $3\times 3$. But is there any way to enforce the positive definiteness of the matrix $A$ via the trace and determinant of $A$, if $A$ is of size $3\times 3$?

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    $\begingroup$ Sylvester's Criterion is the most natural extension of this. Note $\det(A) > 0$ implies both eigenvalues are of the same sign, while $\operatorname{tr}(A) > 0$ implies they must both be positive, or equivalently, the top left $1 \times 1$ matrix has positive determinant. This is Sylvester's Criterion for $2 \times 2$ matrices. Note: you also need to assume that the matrix is Hermitian! $\endgroup$ May 16, 2022 at 15:42

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It is possible to consider the Routh-Hurwitz criterion: A matrix $A$ is Hurwitz stable if its eigenvalues have negative real part. Moreover, a $3\times3$ matrix $A$ with characteristic polynomial $$p_A(x)=x^3+p_2x^2+p_1x+p_0$$

is Hurwitz stable if and only if $p_2>0$, $p_0>0$ and $p_0-p_1p_2<0$.

So, a symmetric matrix $A$ is positive definite if and only if $-A$ is Hurwitz stable.

The characteristic polynomial of $A$ is given by

$$p_A(x)=x^3-\mathrm{trace}(A)x^2+\mathrm{trace}(\mathrm{Adj}(A))x-\det(A)$$

where $\mathrm{Adj}(A)$ is the adjugate matrix of $A$. Consequently, the characteristic polynomial of $-A$ is given by

$$p_{-A}(x)=x^3+\mathrm{trace}(A)x^2-\mathrm{trace}(\mathrm{Adj}(A))x+\det(A).$$

The Routh-Hurwitz criterion for the stability of $-M$ is given by

  • $\mathrm{trace}(A)>0$,
  • $\det(A)>0$,
  • $\det(A)+\mathrm{trace}(\mathrm{Adj}(A))\mathrm{trace}(A)>0$.
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    $\begingroup$ N.b. that $\operatorname{trace}(\operatorname{Adj}(-A)) = \operatorname{trace}(\operatorname{Adj}(A))$. $\endgroup$ May 16, 2022 at 16:24
  • $\begingroup$ @TravisWillse Thanks, I will update the answer. $\endgroup$
    – KBS
    May 16, 2022 at 16:33
  • $\begingroup$ Incidentally, I think we can simplify the criterion using the fact that, a priori, the eigenvalues of a symmetric matrix are real; see my answer. $\endgroup$ May 16, 2022 at 16:56
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No, it is not possible. Take any two real numbers $a$ and $b$ such that $a^2+b^2=1$ and the matrix $$ A= \begin{pmatrix} 1 & a & b \\ a & a^2+1 & a b \\ b & a b & -a^2+2 \end{pmatrix}. $$ Then $\operatorname{tr}(A)=4$, $\det(A)=1$, and $A$ is definite positive.

But if you take two real numbers $a$ and $b$ such that $a^2+b^2=5$ and you define $$ B= \begin{pmatrix} 1&a&b\\ a&a^2-1&ab\\ b&ab&-a^2+4 \end{pmatrix}, $$ then $\operatorname{tr}(B)=4$, $\det(B)=1$, but $B$ is not definite positive.

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    $\begingroup$ How did you find these examples? $\endgroup$ May 16, 2022 at 17:05
  • $\begingroup$ I took a $3\times3$ symmetric matrix$$M=\begin{pmatrix}1&a&b\\a&c&d\\b&c&e\end{pmatrix}$$and then I found a solution of the system$$\left\{\begin{array}{l}\operatorname{tr}(M)=4\\c-a^2=1\\\det(M)=1\end{array}\right.$$That's how I got $A$. In order to get $B$, I did almost the same thing, the only difference being that I used the equation $c-a^2=-1$ instead. $\endgroup$ May 16, 2022 at 17:11
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No, positive definiteness of a (symmetric) $3 \times 3$ matrix $A$ cannot be determined using $\det A$ and $\operatorname{tr} A$ alone.

(Counter)example The diagonal matrix $$\pmatrix{-\frac{1}{2}\\&-\frac{1}{2}\\&&4}$$ has the same trace ($3$) and determinant ($1$) as the (positive definite) $3 \times 3$ identity matrix but is not positive definite.


In any case we can generalize the criterion for a $2 \times 2$ matrix to $3 \times 3$---in fact $n \times n$---matrices.

Denote the eigenvalues $A$ by $\lambda, \mu, \nu$; since $A$ is symmetric, all $3$ are real. The characteristic polynomial of $A$ is $$p_A(t) = t^3 - (\lambda + \mu + \nu) t^2 + (\mu \nu + \nu \lambda + \lambda \mu) t - \lambda \mu \nu,$$ or, more compactly, $$p_A(t) = t^3 - \operatorname{tr} A \cdot t^2 + \sigma_2(A) t - \det A ,$$ where $\sigma_2(A) := \mu \nu + \nu \lambda + \lambda \mu$, i.e., the second elementary symmetric polynomial in the eigenvalues of $A$. So, if $\operatorname{tr} A, \sigma_2(A), \det A > 0$, Descartes' Rule of Signs implies that the (again, real) roots $\lambda, \mu, \nu$ of $p_A$ are positive, equivalently that $A$ is positive definite. Conversely, if $A$ is positive definite, $\sigma_2(A) = \mu \nu + \nu \lambda + \lambda \mu > 0$. Thus, $$\boxed{\textrm{$A$ is positive definite} \qquad \textrm{iff} \qquad \left\{\begin{array}{r} \operatorname{tr} A > 0 \\ \sigma_2(A) > 0 \\ \operatorname{det} A > 0 \end{array}\right.} .$$ The same reasoning immediately yields a generalization to (symmetric) matrices of arbitrary size: $$\boxed{\textrm{$A$ is positive definite} \qquad \textrm{iff} \qquad (\sigma_k(A) = 0 \qquad \textrm{for all } k \in \{1, \ldots, n\})} ,$$ where $\sigma_k(A)$ is the $i$th elementary symmetric polynomial in the eigenvalues $\lambda_1, \ldots, \lambda_n$ of $A$, $$\sigma_k(A) := \sum_{1 \leq i_1 < \cdots i_k \leq n} \lambda_{i_1} \cdots \lambda_{i_k} .$$ Notice that $\sigma_1 = \operatorname{tr}$ and $\sigma_n = \operatorname{det}$, so this statement specializes to the result for $n = 2$ given in the question statement.

Remark We have the identity $\sigma_2 = \operatorname{tr} \circ \operatorname{Adj}$, where $\operatorname{Adj} A$ denotes the adjugate (a.k.a., confusingly, the classical adjoint) of $A$. We can also express $\sigma(A)$ in terms of traces of $A, A^2$: $$\sigma_2(A) = \frac{1}{2}\left[(\operatorname{tr} A)^2 - \operatorname{tr} (A^2)\right].$$

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  • $\begingroup$ I upvoted but I still find it too complex. $\endgroup$
    – Asinomás
    May 16, 2022 at 17:00
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    $\begingroup$ Actually nvm this is great, I don't know if using Rule of signs is necessary but w/e . It might also be good mentioning how the dim $2$ analogue is basically just doing this but easier. $\endgroup$
    – Asinomás
    May 16, 2022 at 17:09
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    $\begingroup$ @Asinomás Thanks for the comment. I added comments to the end of the answer explaining more explicitly how the generalization to the $n \times n$ case works and that the general method specializes to the result OP mentioned for $n = 2$. $\endgroup$ May 16, 2022 at 17:23
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    $\begingroup$ And you're right, we don't need the full strength of Descartes' Rule of Signs; it just shortens the presentation some. All we need is that the alternation of the signs of $p_A(t)$ means that the coefficients of $p_A(-t)$ all have the same sign, hence $p_A(-t)$ has no positive roots, or equivalently that $p_A$ has no negative roots. $\endgroup$ May 16, 2022 at 17:28

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